For a voltaic cell consisting of chromium, an electrode dipped in a 1.20 M chromium (III) nitrate solution and a tin electrode dipped in a 0.400 M tin (II) nitrate solution, the cell potential at 298 K is mathematically given as
Ecell = 0.577 V
<h3 /><h3>What is the cell potential at 298 K?</h3>
Generally, the equation for the Oxidation and Reduction is mathematically given as
Cr(s) ------------------ Cr+3(aq) + 3e- ] x 2 ...O
Sn+2(aq) + 2e- ------------ Sn(s) ] x 3 ...R
Reaction
2 Cr(s) + 3 Sn+2(aq) --------------- 2 Cr+3(aq) + 3 Sn(s)
Therefore
Eicell = - 0.14 - ( - 0.74)
Eicell = 0.60
In conclusion
![Ecell= E0cell - \frac{0.0591}{n} * \frac{log[Cr+3]^2}{ [ Sn+2]^3}](https://tex.z-dn.net/?f=Ecell%3D%20E0cell%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%20%2A%20%5Cfrac%7Blog%5BCr%2B3%5D%5E2%7D%7B%20%5B%20Sn%2B2%5D%5E3%7D)
Ecell = 0.577 V
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Answer:
4,1,5,3,2 (from left to right)
The answer is basic solution
HOPE THIS HELPED
Answer:
The answer is decreased temperature and increased salinity
Explanation:
It is what is known as the thermohaline circulation
The thermohaline circulation moves the water slowly. This water moves mainly due to differences in its relative density. Much denser water sinks over water that is less dense. Two factors impact the density of seawater: temperature and salinity.
Cold water is denser than hot water:
-Water cools when it loses heat, it occurs at high latitudes.
-Water is heated when it receives energy from the sun, at low latitudes.
Saltier water is much denser than water that has less salt:
-Sea water becomes salty if the evaporation rate increases.
-Sea water becomes less salty if there is a water inlet over the sea.
Answer:
Mole fraction of C₄H₄S = 0.55
Explanation:
Mole fraction is moles of solute / Total moles
Total moles are the sum of moles of solute + moles of solvent.
Let's find out the moles of our solute and our solvent.
Mass of solute: 55g
Mass of solvent: 65g
Mol = Mass / molar mass
55 g / 84.06 g/mol = 0.654 moles of C₄H₄S
65 g /123 g/mol = 0.529 moles of C₂H₃BrO
Total moles = 0.654 + 0.529 = 1.183 moles
Mole fraction of thiophene = Moles of tiophene / Total moles
0.654 / 1.183 = 0.55
